Find the minimum value of f(x) = x^2 - 6x + 9.

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Prepare for the AMSOC 26-003 Module A Test. Utilize flashcards and multiple choice questions with hints and explanations. Ace your exam!

Multiple Choice

Find the minimum value of f(x) = x^2 - 6x + 9.

Explanation:
A quadratic’s smallest value occurs at its vertex, and this one can be rewritten as a perfect square. x^2 - 6x + 9 = (x - 3)^2. A square is always greater than or equal to zero, with equality when x = 3. So the smallest value happens at x = 3, giving f(3) = 0. The minimum value is 0.

A quadratic’s smallest value occurs at its vertex, and this one can be rewritten as a perfect square. x^2 - 6x + 9 = (x - 3)^2. A square is always greater than or equal to zero, with equality when x = 3. So the smallest value happens at x = 3, giving f(3) = 0. The minimum value is 0.

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